A convex lens of focal length $15\mathrm{cm}$, is forming a real image. If the size of image is same as the size of object, then position of object and position of image will be, respectively:

As per the NCERT text (Table 9.4, Section 9.3.4), a convex lens forms a real, inverted image of the same size as the object only when the object is placed at $2F_1$; the image is then formed at $2F_2$ on the other side. This can be verified using the lens formula $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$ (Eq. 9.8) and magnification $m=\dfrac{v}{u}$ (Eq. 9.10) from Section 9.3.7.

Given $f = +15\mathrm{cm}$ (positive by the New Cartesian Sign Convention for a convex lens, Section 9.3.6). For the image to be the same size as the object, $|m| = 1$, i.e. $v = -u$ (real image, so $m$ is negative since the object is upright and image is inverted).

Substituting $v = -u$ in the lens formula:

$$\frac{1}{-u} - \frac{1}{u} = \frac{1}{15} \implies \frac{-2}{u} = \frac{1}{15} \implies u = -30\mathrm{cm}$$

Then $v = -u = +30\mathrm{cm}$.

So the object is at $u=-30\mathrm{cm}$ (i.e. at $2F_1$, since $2f = 30\mathrm{cm}$) and the image is at $v=+30\mathrm{cm}$ (at $2F_2$), matching option (C).

Why the other options are wrong:

• (A) $-15\mathrm{cm}$ and $-15\mathrm{cm}$: object at $F_1$ gives an image at infinity (Table 9.4), not a finite same-size image; also both distances cannot be negative for a real image formed on the far side of a convex lens.
• (B) $-15\mathrm{cm}$ and $+15\mathrm{cm}$: object at $F_1$ means rays emerge parallel and no finite image forms at all (Table 9.4, row 5).
• (D) $-30\mathrm{cm}$ and $-30\mathrm{cm}$: a real image formed by a convex lens must be on the opposite side of the lens from the object, so $v$ must be positive (+30 cm), not negative.